Answer:
[tex]P_2=404 kPa[/tex]
Explanation:
Hello,
In this case, the Boyle's is mathematically defined via:
[tex]P_1V_1=P_2V_2[/tex]
Which stands for an inversely proportional relationship between volume and pressure, it means the higher the volume the lower the pressure and vice versa. In such a way, since the volume is decreased to one quarter, we can write:
[tex]V_2=\frac{1}{4} V_1[/tex]
We can compute the new pressure:
[tex]P_2=\frac{P_1V_1}{V_2} =\frac{P_1V_1}{\frac{1}{4} V_1} =\frac{101kPa*V_1}{\frac{1}{4} V_1} \\\\P_2=4*101kPa\\\\\\P_2=404 kPa[/tex]
Which means the pressure is increased by a factor of four.
Regards.
